I am using GCC 6.3 and to my surprise the following code fragment did compile.
auto foo(auto x)
{
return 2.0*x;
}
....
foo(5);
AFAIK it is GCC extension. Compare to the following:
template <typename T, typename R>
R foo(T x)
{
return 2.0*x;
}
Besides return type deduction are the above declaration equivalent?
Using the same GCC (6.3) with the -Wpedantic
flag will generate the following warning:
warning: ISO C++ forbids use of 'auto' in parameter declaration [-Wpedantic]
auto foo(auto x)
^~~~
While compiling this in newer versions of GCC , even without -Wpedantic
, will generate this warning, reminding you about the -fconcepts
flag:
warning: use of 'auto' in parameter declaration only available with -fconcepts
auto foo(auto x)
^~~~
Compiler returned: 0
And indeed, concepts make this:
void foo(auto x)
{
auto y = 2.0*x;
}
equivalent to this:
template<class T>
void foo(T x)
{
auto y = 2.0*x;
}
See here : "If any of the function parameters uses a placeholder (either auto
or a constrained type), the function declaration is instead an abbreviated function template declaration : [...] (concepts TS)" -- emphasis mine.
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